Problem:
$ \int\limits_{-\pi}^{\pi}{\frac{\cos\theta}{a + b \cos\theta}d\theta} $
Solution:
This exercise use the substitution $ z = e^{i\theta} $.
That led to $ dz = ie^{i\theta}d\theta $, and therefore $ \frac{dz}{iz} = d\theta $.
The integration limit becomes a unit circle.
Using the Euler's identity $ \cos \theta + i\sin\theta = e^{i\theta} = z $, we can write trigonometric functions in terms of $ z $
$ \cos \theta = \frac{e^{i\theta} + e^{-i\theta}}{2} = \frac{z + \frac{1}{z}}{2} $.
$ \sin \theta = \frac{e^{i\theta} - e^{-i\theta}}{2i} = \frac{z - \frac{1}{z}}{2i} $.
Here is one more convenient result for factorizing general quadratic equation:
$ \begin{eqnarray*} & & ax^2 + bx + c \\ &=& \frac{1}{4a}(4ax^2 + 4abx + 4ac) \\ &=& \frac{1}{4a}((2ax)^2 + 4abx + 4ac) \\ &=& \frac{1}{4a}((2ax)^2 + 4abx + b^2 - b^2 + 4ac) \\ &=& \frac{1}{4a}((2ax + b)^2 - b^2 + 4ac) \\ &=& \frac{1}{4a}((2ax + b)^2 - (b^2 - 4ac)) \\ &=& \frac{1}{4a}((2ax + b)^2 - (\Delta)^2) \\ &=& \frac{1}{4a}((2ax + b + \Delta)(2ax + b - \Delta)) \end{eqnarray*} $
Now, we are all set to start solving the problems.
$ \begin{eqnarray*} & & \int\limits_{-\pi}^{\pi}{\frac{\cos\theta}{a + b \cos\theta}d\theta} \\ &=& \oint{\frac{\frac{z + \frac{1}{z}}{2}}{a + b \frac{z + \frac{1}{z}}{2}}\frac{dz}{iz}} \\ &=& \oint{\frac{z + \frac{1}{z}}{2a + b (z + \frac{1}{z})}\frac{dz}{iz}} \\ &=& \oint{\frac{z^2 + 1}{2az + b (z^2 + 1)}\frac{dz}{iz}} \\ &=& \oint{\frac{z^2 + 1}{bz^2 + 2az + b}\frac{dz}{iz}} \\ \end{eqnarray*} $
To move on, we factorize the quadratic expression, computing $ \Delta $.
$ \begin{eqnarray*} \Delta &=& \sqrt{(2a)^2 - 4(b)(b)} \\ &=& 2 \sqrt{a^2 - b^2} \end{eqnarray*} $
So we move on
$ \begin{eqnarray*} & & \int\limits_{-\pi}^{\pi}{\frac{\cos\theta}{a + b \cos\theta}d\theta} \\ &=& \oint{\frac{z^2 + 1}{\frac{1}{4b}(2bz + 2a + 2 \sqrt{a^2 - b^2})(2bz + 2a - 2 \sqrt{a^2 - b^2})}\frac{dz}{iz}} \\ &=& \oint{\frac{z^2 + 1}{\frac{1}{b}(bz + a + \sqrt{a^2 - b^2})(bz + a - \sqrt{a^2 - b^2})}\frac{dz}{iz}} \\ &=& \oint{\frac{b(z^2 + 1)}{(bz + a + \sqrt{a^2 - b^2})(bz + a - \sqrt{a^2 - b^2})}\frac{dz}{iz}} \\ \end{eqnarray*} $
We have three poles with 1st order: $ z = 0 $, $ z = -\frac{a + \sqrt{a^2 - b^2}}{b} $ and $ z = \frac{\sqrt{a^2 - b^2} - a}{b} $ Only the first and last is within the unit circle.
Next, we compute the residue values using limits
$ \begin{eqnarray*} & & Res[\frac{b(z^2 + 1)}{(bz + a + \sqrt{a^2 - b^2})(bz + a - \sqrt{a^2 - b^2})}\frac{1}{iz}, 0] \\ &=& \lim\limits_{z \to 0}(z \frac{b(z^2 + 1)}{(bz + a + \sqrt{a^2 - b^2})(bz + a - \sqrt{a^2 - b^2})}\frac{1}{iz}) \\ &=& \lim\limits_{z \to 0}(\frac{b(z^2 + 1)}{(bz + a + \sqrt{a^2 - b^2})(bz + a - \sqrt{a^2 - b^2})}\frac{1}{i}) \\ &=& \frac{b((0)^2 + 1)}{(b(0) + a + \sqrt{a^2 - b^2})(b(0) + a - \sqrt{a^2 - b^2})}\frac{1}{i} \\ &=& \frac{b}{(a + \sqrt{a^2 - b^2})(a - \sqrt{a^2 - b^2})}\frac{1}{i} \\ &=& \frac{b}{(a^2 - (a^2 - b^2))}\frac{1}{i} \\ &=& \frac{b}{b^2}\frac{1}{i} \\ &=& \frac{1}{b}\frac{1}{i} \\ \end{eqnarray*} $
We also have
$ \begin{eqnarray*} & & Res[\frac{b(z^2 + 1)}{(bz + a + \sqrt{a^2 - b^2})(bz + a - \sqrt{a^2 - b^2})}\frac{1}{iz}, \frac{\sqrt{a^2 - b^2} - a}{b}] \\ &=& \lim\limits_{z \to \frac{\sqrt{a^2 - b^2} - a}{b}}((z - \frac{\sqrt{a^2 - b^2} - a}{b}) \frac{b(z^2 + 1)}{(bz + a + \sqrt{a^2 - b^2})(bz + a - \sqrt{a^2 - b^2})}\frac{1}{iz}) \\ &=& \lim\limits_{z \to \frac{\sqrt{a^2 - b^2} - a}{b}}(\frac{1}{b}(bz - (\sqrt{a^2 - b^2} - a)) \frac{b(z^2 + 1)}{(bz + a + \sqrt{a^2 - b^2})(bz + a - \sqrt{a^2 - b^2})}\frac{1}{iz}) \\ &=& \lim\limits_{z \to \frac{\sqrt{a^2 - b^2} - a}{b}}(\frac{1}{b}(bz + a - \sqrt{a^2 - b^2}) \frac{b(z^2 + 1)}{(bz + a + \sqrt{a^2 - b^2})(bz + a - \sqrt{a^2 - b^2})}\frac{1}{iz}) \\ &=& \lim\limits_{z \to \frac{\sqrt{a^2 - b^2} - a}{b}}(\frac{1}{b} \frac{b(z^2 + 1)}{bz + a + \sqrt{a^2 - b^2}}\frac{1}{iz}) \\ &=& \frac{1}{b} \frac{b((\frac{\sqrt{a^2 - b^2} - a}{b})^2 + 1)}{b(\frac{\sqrt{a^2 - b^2} - a}{b}) + a + \sqrt{a^2 - b^2}}\frac{1}{i(\frac{\sqrt{a^2 - b^2} - a}{b})} \\ &=& \frac{1}{b} \frac{b((\frac{\sqrt{a^2 - b^2} - a}{b})^2 + 1)}{\sqrt{a^2 - b^2} - a + a + \sqrt{a^2 - b^2}}\frac{1}{i(\frac{\sqrt{a^2 - b^2} - a}{b})} \\ &=& \frac{1}{b} \frac{b((\frac{\sqrt{a^2 - b^2} - a}{b})^2 + 1)}{2\sqrt{a^2 - b^2}}\frac{1}{i(\frac{\sqrt{a^2 - b^2} - a}{b})} \\ &=& \frac{1}{b} \frac{b^2((\frac{\sqrt{a^2 - b^2} - a}{b})^2 + 1)}{2\sqrt{a^2 - b^2}}\frac{1}{i(\sqrt{a^2 - b^2} - a)} \\ &=& \frac{1}{b} \frac{(\sqrt{a^2 - b^2} - a)^2 + b^2}{2\sqrt{a^2 - b^2}}\frac{1}{i(\sqrt{a^2 - b^2} - a)} \\ &=& \frac{1}{b} \frac{(a^2 - b^2) - 2a\sqrt{a^2 - b^2} + a^2 + b^2}{2\sqrt{a^2 - b^2}}\frac{1}{i(\sqrt{a^2 - b^2} - a)} \\ &=& \frac{1}{b} \frac{2a^2 - 2a\sqrt{a^2 - b^2}}{2\sqrt{a^2 - b^2}}\frac{1}{i(\sqrt{a^2 - b^2} - a)} \\ &=& \frac{1}{b} \frac{-2a(\sqrt{a^2 - b^2} - a)}{2\sqrt{a^2 - b^2}}\frac{1}{i(\sqrt{a^2 - b^2} - a)} \\ &=& \frac{1}{b} \frac{-a}{\sqrt{a^2 - b^2}}\frac{1}{i} \\ \end{eqnarray*} $
Last, we put these back to the integral, we have
$ \begin{eqnarray*} & & \int\limits_{-\pi}^{\pi}{\frac{\cos\theta}{a + b \cos\theta}d\theta} \\ &=& 2\pi i(\frac{1}{b}\frac{1}{i} + \frac{1}{b} \frac{-a}{\sqrt{a^2 - b^2}}\frac{1}{i}) \\ &=& \frac{2\pi}{b} (1 - \frac{a}{\sqrt{a^2 - b^2}}) \\ \end{eqnarray*} $
Q.E.D. This took me 1.5 hours.
$ \int\limits_{-\pi}^{\pi}{\frac{\cos\theta}{a + b \cos\theta}d\theta} $
Solution:
This exercise use the substitution $ z = e^{i\theta} $.
That led to $ dz = ie^{i\theta}d\theta $, and therefore $ \frac{dz}{iz} = d\theta $.
The integration limit becomes a unit circle.
Using the Euler's identity $ \cos \theta + i\sin\theta = e^{i\theta} = z $, we can write trigonometric functions in terms of $ z $
$ \cos \theta = \frac{e^{i\theta} + e^{-i\theta}}{2} = \frac{z + \frac{1}{z}}{2} $.
$ \sin \theta = \frac{e^{i\theta} - e^{-i\theta}}{2i} = \frac{z - \frac{1}{z}}{2i} $.
Here is one more convenient result for factorizing general quadratic equation:
$ \begin{eqnarray*} & & ax^2 + bx + c \\ &=& \frac{1}{4a}(4ax^2 + 4abx + 4ac) \\ &=& \frac{1}{4a}((2ax)^2 + 4abx + 4ac) \\ &=& \frac{1}{4a}((2ax)^2 + 4abx + b^2 - b^2 + 4ac) \\ &=& \frac{1}{4a}((2ax + b)^2 - b^2 + 4ac) \\ &=& \frac{1}{4a}((2ax + b)^2 - (b^2 - 4ac)) \\ &=& \frac{1}{4a}((2ax + b)^2 - (\Delta)^2) \\ &=& \frac{1}{4a}((2ax + b + \Delta)(2ax + b - \Delta)) \end{eqnarray*} $
Now, we are all set to start solving the problems.
$ \begin{eqnarray*} & & \int\limits_{-\pi}^{\pi}{\frac{\cos\theta}{a + b \cos\theta}d\theta} \\ &=& \oint{\frac{\frac{z + \frac{1}{z}}{2}}{a + b \frac{z + \frac{1}{z}}{2}}\frac{dz}{iz}} \\ &=& \oint{\frac{z + \frac{1}{z}}{2a + b (z + \frac{1}{z})}\frac{dz}{iz}} \\ &=& \oint{\frac{z^2 + 1}{2az + b (z^2 + 1)}\frac{dz}{iz}} \\ &=& \oint{\frac{z^2 + 1}{bz^2 + 2az + b}\frac{dz}{iz}} \\ \end{eqnarray*} $
To move on, we factorize the quadratic expression, computing $ \Delta $.
$ \begin{eqnarray*} \Delta &=& \sqrt{(2a)^2 - 4(b)(b)} \\ &=& 2 \sqrt{a^2 - b^2} \end{eqnarray*} $
So we move on
$ \begin{eqnarray*} & & \int\limits_{-\pi}^{\pi}{\frac{\cos\theta}{a + b \cos\theta}d\theta} \\ &=& \oint{\frac{z^2 + 1}{\frac{1}{4b}(2bz + 2a + 2 \sqrt{a^2 - b^2})(2bz + 2a - 2 \sqrt{a^2 - b^2})}\frac{dz}{iz}} \\ &=& \oint{\frac{z^2 + 1}{\frac{1}{b}(bz + a + \sqrt{a^2 - b^2})(bz + a - \sqrt{a^2 - b^2})}\frac{dz}{iz}} \\ &=& \oint{\frac{b(z^2 + 1)}{(bz + a + \sqrt{a^2 - b^2})(bz + a - \sqrt{a^2 - b^2})}\frac{dz}{iz}} \\ \end{eqnarray*} $
We have three poles with 1st order: $ z = 0 $, $ z = -\frac{a + \sqrt{a^2 - b^2}}{b} $ and $ z = \frac{\sqrt{a^2 - b^2} - a}{b} $ Only the first and last is within the unit circle.
Next, we compute the residue values using limits
$ \begin{eqnarray*} & & Res[\frac{b(z^2 + 1)}{(bz + a + \sqrt{a^2 - b^2})(bz + a - \sqrt{a^2 - b^2})}\frac{1}{iz}, 0] \\ &=& \lim\limits_{z \to 0}(z \frac{b(z^2 + 1)}{(bz + a + \sqrt{a^2 - b^2})(bz + a - \sqrt{a^2 - b^2})}\frac{1}{iz}) \\ &=& \lim\limits_{z \to 0}(\frac{b(z^2 + 1)}{(bz + a + \sqrt{a^2 - b^2})(bz + a - \sqrt{a^2 - b^2})}\frac{1}{i}) \\ &=& \frac{b((0)^2 + 1)}{(b(0) + a + \sqrt{a^2 - b^2})(b(0) + a - \sqrt{a^2 - b^2})}\frac{1}{i} \\ &=& \frac{b}{(a + \sqrt{a^2 - b^2})(a - \sqrt{a^2 - b^2})}\frac{1}{i} \\ &=& \frac{b}{(a^2 - (a^2 - b^2))}\frac{1}{i} \\ &=& \frac{b}{b^2}\frac{1}{i} \\ &=& \frac{1}{b}\frac{1}{i} \\ \end{eqnarray*} $
We also have
$ \begin{eqnarray*} & & Res[\frac{b(z^2 + 1)}{(bz + a + \sqrt{a^2 - b^2})(bz + a - \sqrt{a^2 - b^2})}\frac{1}{iz}, \frac{\sqrt{a^2 - b^2} - a}{b}] \\ &=& \lim\limits_{z \to \frac{\sqrt{a^2 - b^2} - a}{b}}((z - \frac{\sqrt{a^2 - b^2} - a}{b}) \frac{b(z^2 + 1)}{(bz + a + \sqrt{a^2 - b^2})(bz + a - \sqrt{a^2 - b^2})}\frac{1}{iz}) \\ &=& \lim\limits_{z \to \frac{\sqrt{a^2 - b^2} - a}{b}}(\frac{1}{b}(bz - (\sqrt{a^2 - b^2} - a)) \frac{b(z^2 + 1)}{(bz + a + \sqrt{a^2 - b^2})(bz + a - \sqrt{a^2 - b^2})}\frac{1}{iz}) \\ &=& \lim\limits_{z \to \frac{\sqrt{a^2 - b^2} - a}{b}}(\frac{1}{b}(bz + a - \sqrt{a^2 - b^2}) \frac{b(z^2 + 1)}{(bz + a + \sqrt{a^2 - b^2})(bz + a - \sqrt{a^2 - b^2})}\frac{1}{iz}) \\ &=& \lim\limits_{z \to \frac{\sqrt{a^2 - b^2} - a}{b}}(\frac{1}{b} \frac{b(z^2 + 1)}{bz + a + \sqrt{a^2 - b^2}}\frac{1}{iz}) \\ &=& \frac{1}{b} \frac{b((\frac{\sqrt{a^2 - b^2} - a}{b})^2 + 1)}{b(\frac{\sqrt{a^2 - b^2} - a}{b}) + a + \sqrt{a^2 - b^2}}\frac{1}{i(\frac{\sqrt{a^2 - b^2} - a}{b})} \\ &=& \frac{1}{b} \frac{b((\frac{\sqrt{a^2 - b^2} - a}{b})^2 + 1)}{\sqrt{a^2 - b^2} - a + a + \sqrt{a^2 - b^2}}\frac{1}{i(\frac{\sqrt{a^2 - b^2} - a}{b})} \\ &=& \frac{1}{b} \frac{b((\frac{\sqrt{a^2 - b^2} - a}{b})^2 + 1)}{2\sqrt{a^2 - b^2}}\frac{1}{i(\frac{\sqrt{a^2 - b^2} - a}{b})} \\ &=& \frac{1}{b} \frac{b^2((\frac{\sqrt{a^2 - b^2} - a}{b})^2 + 1)}{2\sqrt{a^2 - b^2}}\frac{1}{i(\sqrt{a^2 - b^2} - a)} \\ &=& \frac{1}{b} \frac{(\sqrt{a^2 - b^2} - a)^2 + b^2}{2\sqrt{a^2 - b^2}}\frac{1}{i(\sqrt{a^2 - b^2} - a)} \\ &=& \frac{1}{b} \frac{(a^2 - b^2) - 2a\sqrt{a^2 - b^2} + a^2 + b^2}{2\sqrt{a^2 - b^2}}\frac{1}{i(\sqrt{a^2 - b^2} - a)} \\ &=& \frac{1}{b} \frac{2a^2 - 2a\sqrt{a^2 - b^2}}{2\sqrt{a^2 - b^2}}\frac{1}{i(\sqrt{a^2 - b^2} - a)} \\ &=& \frac{1}{b} \frac{-2a(\sqrt{a^2 - b^2} - a)}{2\sqrt{a^2 - b^2}}\frac{1}{i(\sqrt{a^2 - b^2} - a)} \\ &=& \frac{1}{b} \frac{-a}{\sqrt{a^2 - b^2}}\frac{1}{i} \\ \end{eqnarray*} $
Last, we put these back to the integral, we have
$ \begin{eqnarray*} & & \int\limits_{-\pi}^{\pi}{\frac{\cos\theta}{a + b \cos\theta}d\theta} \\ &=& 2\pi i(\frac{1}{b}\frac{1}{i} + \frac{1}{b} \frac{-a}{\sqrt{a^2 - b^2}}\frac{1}{i}) \\ &=& \frac{2\pi}{b} (1 - \frac{a}{\sqrt{a^2 - b^2}}) \\ \end{eqnarray*} $
Q.E.D. This took me 1.5 hours.
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